SDKD 2016 Summer Single Contest #01.E

Pari wants to buy an expensive chocolate from Arya. She has n coins, the value of the i-th coin is ci. The price of the chocolate is k, so Pari will take a subset of her coins with sum equal to k and give it to Arya.

Looking at her coins, a question came to her mind: after giving the coins to Arya, what values does Arya can make with them? She is jealous and she doesn’t want Arya to make a lot of values. So she wants to know all the values x, such that Arya will be able to make x using some subset of coins with the sum k.

Formally, Pari wants to know the values x such that there exists a subset of coins with the sum k such that some subset of this subset has the sum x, i.e. there is exists some way to pay for the chocolate, such that Arya will be able to make the sum x using these coins.

Input

The first line contains two integers n and k (1  ≤  n, k  ≤  500) — the number of coins and the price of the chocolate, respectively.

Next line will contain n integers c1, c2, …, cn (1 ≤ ci ≤ 500) — the values of Pari’s coins.

It’s guaranteed that one can make value k using these coins.

Output

First line of the output must contain a single integer q— the number of suitable values x. Then print q integers in ascending order — the values that Arya can make for some subset of coins of Pari that pays for the chocolate.

Sample Input

Input

6 18

5 6 1 10 12 2

Output

16

0 1 2 3 5 6 7 8 10 11 12 13 15 16 17 18

Input

3 50

25 25 50

Output

3

0 25 50

题目:

dp是一点儿也不会,只能是慢慢的懂这个代码的意思了,dp的题还是做不了~,

这是队里大牛的代码,他用的是vector 我直接换了set.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
const int maxn = 500 + 50;
int n, k, m;
int dp[maxn][maxn];
set<int >s;
int main()
{
cin >> n >> k;
s.clear();
memset(dp, 0, sizeof(dp));
dp[0][0] = 1;
for(int i = 0; i < n; i++)
{
scanf("%d", &m);
//j 代表 当前 的和;
for(int j = k; j >= m; j--)
{
//l代表子集里的数;
for(int l = 0; l + m <= k; l++)
{
//如果[j-m][l] = 1的话,说明之前的
// [a][b] = 1;a = j-m,b = l;
//[a][b]的意思是 和a 有子集 b,
//那么a+m  肯定也有 子集 b; 因为 a+m > a;
//那么 a+m 也有子集 b+m;
//具体的我也搞不懂了,dp还要在研究;
if(dp[j-m][l] == 1)
dp[j][l] = dp[j][l + m] = 1;
}
}
}
for(int i = 0; i <= k; i++)
if(dp[k][i] != 0)
s.insert(i);
printf("%d\n", s.size());
for(set<int > :: iterator it = s.begin(); it != s.end(); it++)
{
if(it != s.begin())
printf(" ");
cout << *it;
}
printf("\n");
return 0;
}